Problem 22 - Calculus Early Transcendentals

Question

\(21-26\) Use the Chain Rule to find the indicated partial derivatives. $$\begin{array}{l}{u=\sqrt{r^{2}+s^{2}}, \quad r=y+x \cos t, \quad s=x+y \sin t} \\ {\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial t} \quad \text { when } x=1, y=2, t=0}\end{array}$$

Step-by-Step Solution

Answer

\( \frac{\partial u}{\partial x} = \frac{4}{\sqrt{10}}, \frac{\partial u}{\partial y} = \frac{3}{\sqrt{10}}, \frac{\partial u}{\partial t} = \frac{2}{\sqrt{10}}. \)

Step 1: Understand the Problem

We need to find the partial derivatives \( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \) and \( \frac{\partial u}{\partial t} \) of the function \( u = \sqrt{r^2 + s^2} \), where \( r = y + x \cos t \) and \( s = x + y \sin t \), evaluated at \( x=1, y=2, \) and \( t=0 \). We will use the Chain Rule for differentiation.

Step 2: Compute \( \frac{\partial u}{\partial x} \) Using the Chain Rule

First, find the partial derivatives \( \frac{\partial r}{\partial x} \) and \( \frac{\partial s}{\partial x} \):\[ \frac{\partial r}{\partial x} = \cos t, \quad \frac{\partial s}{\partial x} = 1. \]Now apply the Chain Rule:\[ \frac{\partial u}{\partial x} = \frac{1}{2\sqrt{r^2 + s^2}} (2r \frac{\partial r}{\partial x} + 2s \frac{\partial s}{\partial x}) = \frac{r \cos t + s}{\sqrt{r^2 + s^2}}. \]Substitute \( x=1, y=2, t=0 \):- \( r = 2 + 1 \cos 0 = 3 \), \( s = 1 + 2 \sin 0 = 1 \), and thus \( \sqrt{r^2 + s^2} = \sqrt{10} \).- \( \frac{\partial u}{\partial x} = \frac{3 \cdot 1 + 1}{\sqrt{10}} = \frac{4}{\sqrt{10}}. \)

Step 3: Compute \( \frac{\partial u}{\partial y} \) Using the Chain Rule

Find the partial derivatives \( \frac{\partial r}{\partial y} \) and \( \frac{\partial s}{\partial y} \):\[ \frac{\partial r}{\partial y} = 1, \quad \frac{\partial s}{\partial y} = \sin t. \]Apply the Chain Rule:\[ \frac{\partial u}{\partial y} = \frac{r\frac{\partial r}{\partial y} + s\frac{\partial s}{\partial y}}{\sqrt{r^2 + s^2}} = \frac{r + s \sin t}{\sqrt{r^2 + s^2}}. \]Insert the values \( x=1, y=2, t=0 \):- With \( r = 3 \) and \( s = 1 \), \( \sin 0 = 0 \), we get:\( \frac{\partial u}{\partial y} = \frac{3 + 1 \cdot 0}{\sqrt{10}} = \frac{3}{\sqrt{10}}. \)

Step 4: Compute \( \frac{\partial u}{\partial t} \) Using the Chain Rule

Determine the partial derivatives \( \frac{\partial r}{\partial t} \) and \( \frac{\partial s}{\partial t} \):\[ \frac{\partial r}{\partial t} = -x \sin t, \quad \frac{\partial s}{\partial t} = y \cos t. \]Apply the Chain Rule:\[ \frac{\partial u}{\partial t} = \frac{r \frac{\partial r}{\partial t} + s \frac{\partial s}{\partial t}}{\sqrt{r^2 + s^2}}. \]Substitute the given values, \( x=1, y=2, t=0 \):- \( \frac{\partial r}{\partial t} = -1 \cdot 0 = 0 \) and \( \frac{\partial s}{\partial t} = 2 \cdot 1 = 2 \).- \( \frac{\partial u}{\partial t} = \frac{3 \cdot 0 + 1 \cdot 2}{\sqrt{10}} = \frac{2}{\sqrt{10}}. \)

Key Concepts

Partial Derivatives

When working with functions of multiple variables, a concept called partial derivatives becomes essential. Partial derivatives measure how a function changes as one specific variable changes, holding all others constant. For example, if you have a function \( u(r, s) \), its partial derivative \( \frac{\partial u}{\partial r} \) looks at how \( u \) changes with \( r \) while keeping \( s \) constant.
In this exercise, we compute partial derivatives of \( u = \sqrt{r^2 + s^2} \) with respect to \( x \), \( y \), and \( t \). These can involve taking the derivatives of intermediate variables, such as \( r \) and \( s \), with respect to \( x \), \( y \), and \( t \).
The calculations involve understanding how each variable influences the function individually, which is a core concept in calculus.

Calculus

Calculus provides a framework for understanding changes and areas of functions, and partial derivatives are just one of the powerful tools it offers. It's divided into two main branches: differentiation and integration. In this context, we focus on differentiation, which examines the rate of change of quantities.

• Differentiation helps us understand how a multivariable function changes when one of its input variables changes.
• It provides insight into how a function's value is sensitive to changes in its variables.

Understanding these changes is crucial in fields like physics and engineering where functions with multiple inputs are common. Calculus, and specifically the Chain Rule, helps us navigate these complex relationships.

Differentiation

Differentiation is the mathematical process of finding a derivative, which is a measure of how a function changes as its input changes. The derivative captures the function's sensitivity to changes in its inputs. In the realm of multivariable calculus, this means finding partial derivatives.
Using the Chain Rule, we compute derivatives when a function is composed of other functions. Consider \( u = \sqrt{r^2 + s^2} \) where \( r \) and \( s \) themselves depend on \( x \), \( y \), and \( t \). The Chain Rule allows us to differentiate \( u \) with respect to these variables, breaking it down through intermediate stages, such as finding \( \frac{\partial r}{\partial x} \) and \( \frac{\partial s}{\partial x} \).
The Chain Rule efficiently guides us through each step, ensuring that we account for all ways inputs affect the output.

Multivariable Functions

Multivariable functions are functions that depend on more than one variable. For example, \( u = \sqrt{r^2 + s^2} \) depends on variables \( x \), \( y \), and \( t \). These functions are common in situations where outcomes depend on several factors, like physics or economics.
To understand these functions fully, it’s often necessary to consider how they change as one variable changes, which involves finding partial derivatives.

• Each partial derivative tells us about the rate of change along one axis or variable.
• Multivariable calculus offers tools like the Chain Rule to differentiate these functions efficiently.

Knowing how to navigate and manipulate multivariable functions opens up a world of solutions in technologically advanced fields where systems are inherently complex.